What is the derivative of 4/(16x^2+1) ?

The derivative of 4/(16x^2+1) is -(128x)/((16x^2+1)^2)

What is the first derivative of 4/(16x^2+1) ?

The first derivative of 4/(16x^2+1) is -(128x)/((16x^2+1)^2)

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\frac{d}{d x} (\frac{4}{16 x ^{2} + 1})

- \frac{128 x}{( 16 x ^{2} + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{4}{16 x ^{2} + 1})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 4 \frac{d}{d x} (\frac{1}{16 x ^{2} + 1})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= 4 \frac{d}{d x} ((16 x^{2} + 1)^{- 1})

Apply the chain rule:

- \frac{1}{( 16 x ^{2} + 1 ) ^{2}} \frac{d}{d x} (16 x^{2} + 1)

= - \frac{1}{( 16 x ^{2} + 1 ) ^{2}} \frac{d}{d x} (16 x^{2} + 1)

\frac{d}{d x} (16 x^{2} + 1) = 32 x

= 4 (- \frac{1}{( 16 x ^{2} + 1 ) ^{2}} \cdot 32 x)

Simplify

4 (- \frac{1}{( 16 x ^{2} + 1 ) ^{2}} \cdot 32 x) : - \frac{128 x}{( 16 x ^{2} + 1 ) ^{2}}

= - \frac{128 x}{( 16 x ^{2} + 1 ) ^{2}}

\frac{\partial}{\partial x} (e^{x y})

y = (\frac{3 x ^{2}}{2 x + 4})^{3}

\frac{5}{s ( s ^{2} + 5 ^{2} )}

\frac{d}{d x} ((y - c)^{2})

\int cos^{4} (5 x + 4) d x

What is the derivative of 4/(16x^2+1) ?
The derivative of 4/(16x^2+1) is -(128x)/((16x^2+1)^2)
What is the first derivative of 4/(16x^2+1) ?
The first derivative of 4/(16x^2+1) is -(128x)/((16x^2+1)^2)